p { margin-bottom: 0.25cm; line-height: 120%; }a:link { }This circuit was born out of a need to ensure that
a shower enclosure extractor fan, which would
normally be controlled from a household lighting circuit,
would always run when the shower was used, regardless of whether the
user had turned on the lights or not, and would not run continuously
if the lights were left on. The shower in question
was pump assisted, so that seemed the place to be looking
for a control signal. The design aim was to allow sensing of when
the pump was running, but without the need to "tee in" to
the circuit in anyway. Hence this non-intrusive sensing circuit,
which has the added advantage of being electrically isolated
from the pump circuit.
How it Works
One leg of the mains feed to the load is threaded through the
centre hole of the current transformer. When the load turns on, an AC
voltage appears across the secondary winding of the transformer. This
voltage is rectified by Schottky diode D1, and smoothed by C1 to
produce a DC voltage which, up to a point, is proportional to the
amount of current being drawn by the load. R3 is included to ensure
that C1 is discharged within a couple of seconds of the pump
switching off.
The DC voltage is next applied to the inputs of a CMOS NAND gate
via R1. Schottky diode D3 shunts the internal static protection
diode, and ensures that the voltage being applied to the gate cannot
exceed rail plus approximately 0.3 volts. R1 limits the clamping
current through the diode.
Gate U1a inverts the C1 voltage, so pin 3 drops close to zero
volts. This point drives gates U1b,c & d, which are also
connected as simple inverters. Thus, pin 4 will rise to near rail,
and LED 1 will illuminate to show that the circuit has
switched. R2 limits the LED current to a value that the gate can cope
with.
Gates U1c and U1d are connected in parallel to double their
current drive capability. This would only be an issue if the relay
was a particularly "meaty" one, and the transistor did not
have a lot of gain. Q1 is the relay driver transistor, R4 limiting
its base current to a safe value. When its base is driven high by the
signal from U1c & d, its collector will drop close to zero volts,
causing the relay to pull in. Diode D2 is needed to clamp the reverse
voltage spike produced by the relay coil, when Q1 turns back off.
Without this diode, the transistor would be damaged.
The relay used has mains rated contacts, and will
typically be located close to the load to be switched (in the case of
the original fan switching application it was placed in the fan
control unit itself with the contacts connected across the "Live"’
and "Switch" contacts. Thus, the relay takes the place of
the light switch connection which would normally go to the fan unit).
Notes on Choice of
Components
The circuit was designed to use components that "came to
hand" and most are not critical. However, that said, in order to
get best sensitivity from the circuit, the specified current
transformer type TA17-04 should be used. This is readily available
from a number of Far East and eBay traders at very
reasonable cost. See for instance, see this item
D1 can be any small signal diode. A BAT85 Schottky was used
because I had some lying about and this type of diode will improve
sensitivity. C1 can be any value in the vicinity of 4u7 but should be
rated at least 35v. R3 can be any value from about 470k up to a
couple of meg. The higher the value, the longer the circuit will take
to switch back off after it stops sensing the load being monitored.
Diode D3 needs to be a Schottky to ensure that it conducts to clamp
the input voltage to a safe level, before the IC's internal "normal"
diode would conduct. If using the circuit to sense very high currents
(say 25A or more), then one could substitute a different Schottky
diode with a higher reverse voltage threshold since the output from
the current transformer may exceed the rated value at high loads
(experiments have shown however the device specified functions
without failure even at very high sensed currents). R1 can be any
value from about 2k2 to 10k.
U1 is a common CD4011 quad 2 input NAND gate, but ensure that you
use the buffered variety, which most that you will buy, are. R2 needs
to be chosen to ensure that the current that the gate is sourcing to
drive the LED, does not exceed about 10mA. With a 5v supply, and
a red HE LED, 560 or 680 ohms should be used. For a 12v supply, use
1k5 or 2k2. For different colour LEDs, a different value may be
needed.
Q1 is any general purpose NPN transistor. The 2N3704 shown is good
for 500mA of collector current so should be capable of driving most
modern relays. It also has very good gain, allowing R4 to be 2k7 or
higher, ensuring that the gates are not overloaded. I would not
recommend the use of a power transistor in this position due to their
poor gain, and high drive current requirements. D2 should be at least
a 1N4004. Others, higher in the same series, are also OK. The relay
coil voltage should be chosen to suit the supply voltage being used.
Notes on safety
This circuit uses a current transformer with a high turns ratio
(2000:1). Current transformers have a bit of a reputation for posing
shock risks in some circumstances (e.g. if their outputs are not
adequately terminated/loaded). Experiments with the specified device
have shown however that the voltage output does not rise above around
14V even at at 3kW sensed load. When monitoring higher loads then
ensure that R5 is fitted and there should be no risk. It is strongly
advised that only the specified current transformer is used. Other
devices may give significantly different results.
Adjusting circuit
sensitivity
As shown powered from a 5V supply, the circuit will sense loads
ranging from around 30W to over 4kW (with a 12V supply the minimum
trigger load will rise a little to around 40W). Note that on very
light loads the circuit may take a few moment to switch the relay.
For reliably sensing low loads, one can fool the current
transformer into seeing a bigger load than is actually present by
looping the wirebeing monitored through it two or three times.
This will double or triple the current that it "sees".
The current transformer used in this design has a theoretical
maximum current sensing capacity of 20A. Attempting to sense larger
current will have two effects. Firstly the output voltage could rise
higher than really wanted. This can be dealt with by adding R5 to
load the transformer a little and reduce its output voltage. The
other effect of exceeding the 20A threshold is that the transformer
begins to saturate and lose its linearity. For applications where one
is attempting to accurately sense the value of the current being
monitored this would matter, however in this application where all we
care about is is it "on" or not, it does not matter.
Power supplies
As previously described, the circuit was designed with cheapness
and using what came to hand in mind. To this end, I chose to use a
regulated 5v DC ‘brick’ that I had lying around from a
long-deceased external hard drive, so the cost of this was nothing.
I’m sure that most people will have redundant mobile phone chargers
lying around, and many of these will be 5v. The actual current
requirement of the circuit is almost nothing, being mainly the
current to drive the LED plus that needed for the relay coil. Any
commercial power supply that you might have surplus, will be fine in
that regard. The circuit has been tested with a range of supply
voltages, and it works without problem from 5v to 12v, so any power
supply – either regulated or unregulated – will be ok to use. The
main proviso is that the output voltage must not under any
circumstances exceed 15v, as this is the upper voltage supply limit
for the IC. If intending to use an unregulated supply with any
declared output voltage of 10v or higher, I would suggest that its
off-load voltage is actually measured first, as some unregulated
supplies can deliver a light-load voltage of 50% more than the marked
value.
Regulation
In the event that you want to use a power supply that delivers in
excess of 15v, but not more than 25v, you can add a regulator to the
output as shown below. This circuit uses only three components – a
cheap three terminal regulator IC and two small capacitors. The value
of these capacitors is not critical, but note that they should be
mounted as close to the IC pins as possible. Either the 1 amp version
of the IC or its little brother, the "L" version, will be
fine. With input voltages at the high end of the range, a small
heatsink may be required.
Or, more likely, I'll make me one, and that'll make me happy. Thanks for any and all suggestions!
